You are a product manager and currently leading a team to develop a new product. Unfortunately, the latest version of your product fails the quality check. Since each version is developed based on the previous version, all the versions after a bad version are also bad.
Suppose you have n versions [1,
2, ..., n] and you want to find out the first bad one, which causes all the following ones to be bad.
You are given an API bool isBadVersion(version) which will return whether version is
bad. Implement a function to find the first bad version. You should minimize the number of calls to the API.
/* The isBadVersion API is defined in the parent class VersionControl.
boolean isBadVersion(int version); */
public class Solution extends VersionControl {
public int firstBadVersion(int n) {
long st = 1, ed = n;
Solution v = new Solution();
while(st < ed) {
int mid = (int)((st+ed)>>1);
System.out.println(mid);
boolean is = v.isBadVersion(mid);
if(!is && v.isBadVersion(mid+1)) {
return mid+1;
} else if(is) {
// x x
ed = mid;
} else {
// v v
st = mid+1;
}
}
return (int)st;
}
}
产品团队在开发新产品过程中遇到质量检查失败的问题。为了找到第一个导致后续版本不良的版本,实现了一个函数来最小化调用API次数。

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